# A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute. Find

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Updated on: September 25th, 2023

(a) the total distance moved by train,

(b) the maximum speed attained by the train, and

(c) the position(s) of the train at half the maximum speed.

(a) the total distance moved by train is 2.7 km,

(b) the maximum speed attained by the train is 60 m/s, and

(c) the position(s) of the train at half the maximum speed is 2.25 km.

Step 1: Given that acceleration, a = 2.0 m/s2

Initial velocity, u = 0

Step 2:

Using the equations of motion we have to calculate the final velocity and the total distance traveled.

Final velocity, v = u + at

Distance traveled, x = ut + ½ at2

Step 3: In case of accelerated motion

Given that, a = 2.0 m/s2

Let v1 represent the final speed before applying breaks.

we know that, initial velocity, u1 = 0

Time taken, t1 = ½ x 60 = 30 seconds

Final velocity is now provided by,

v1 = u1 + at1

Substituting the values we get

v1 = 0 + 2 x 30

v1 = 60 m/s

The top speed reached after applying brakes is 60 m/s2

Distance traveled

x = u1t1 + ½ at12

x = 0 + ½ x 2 x 302

x = 900 m

x = 0.9 km

The maximum speed attained is 60 m/s2, Let y be the distance covered while moving at half the maximum speed of 30 m/s.

From the equation now

v12 = u12 + 2ax

302 = 00 + 2 x 2 x y

y = 900/4 = 225 m ….. (1)

As a result, the starting point is 225 metres away.

Step 4: In case of decelerated motion

Final velocity, v2 = 0 m/s

Initial velocity, u2 = 60 m/s

Time taken, t2 = 60 s

From the equation

v2 = u2 + at2

a = (0-60)/60 = -1 m/s2

Distance traveled,

x = u2t2 + ½ at22

Substituting the values we get

x = 60 x 60 + ½ x (-1) x 602

x = 3600 – 1800

x = 1800 m

x = 1.8 km

The total distance covered = 0.9 + 1.8 = 2.7 km

Let z represent the distance travelled in this portion at half the speed.

From

v22 = u22 + 2az

302 = 602 + 2 x (-1) x z

2z = 602 – 302

2z = 2700

z = 1350m

z = 1.35 km …. (2)

Step 5: Conclusion

(a) Total distance traveled by train is given as,

s = x + z

= 900 + 1800

= 2700 m

= 2.7 km

(b) The train can travel at a maximum speed of 60 m/s.

(c) The position of the train is indicated at a speed of half the maximum,

Two places will operate at half the maximum speed (From the point of start),

The position is 225 metres from the starting point.

After braking, the position is = 1.35 km + 0.9 km = 2.25 km

Summary:-

## A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute.

(a) the total distance moved by train is 2.7 km,

(b) the maximum speed attained by the train is 60 m/s, and

(c) the position(s) of the train at half the maximum speed is 2.25 km.

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